Problem Statement:

Let $X_1$ and $X_2$ be independent random variables.

Given the density function, $ f_U(u)$, and the relationship:


Can we find the density functions for $X_1$ and $X_2$?


I know I can use the transformation method to find the conditional density and then use method of conditioning to find $f_U(u)$ given $f_{X_1}(x_1)$ and $f_{X_2}(x_2)$:

$$f(u|x_2)=f_{x_1}(ux_2)∣\frac{dx_1}{du}∣$$ $$f_U(u)=\int_{-\infty}^\infty f(u|x_2)f(x_2)dx_2$$

However, I am wondering if there is any method for reversing this procedure.

Problem Context:

I have 2 uniform random variables and I calculate the distribution of U as described above. I truncate a section of the U distribution and replace it with a delta function. I would like to calculate the distribution of $X_1$ and $X_2$ that correspond to the new U distribution.



Given the distribution of the random variable $U=X_1/X_2$ and the hypothesis that $X_1$ and $X_2$ are independent, can we find the distributions of $X_1$ and $X_2$?

Not always.

To see why, assume that $U=X_1/X_2$ for some independent $X_1$ and $X_2$ then $U=Y_1/Y_2$ where $Y_1=1/X_2$ and $Y_2=1/X_1$ are independent. If the distributions of $X_1$ and $1/X_2$ do not coincide, this provides another couple of distributions which is solution.

Note that if $X_1$ and $X_2$ are i.i.d. and if the distributions of $X_1$ and $1/X_1$ do not coincide, the new solution $(Y_1,Y_2)$ is also i.i.d. and uses a different distribution.

Still more elementary: for every nondegenerate random variable $U$ such that $P[U=0]=0$, two different solutions are $(X_1,X_2)=(U,1)$ and $(X_1,X_2)=(1,1/U)$.

  • $\begingroup$ I understand your explanation that this is not always possible. In the context of this problem, is it possible and if so what method can I apply? $\endgroup$ – Nat Sep 12 '13 at 2:16
  • $\begingroup$ I do not understand the description of the problem context in your post. Why should we suspect that the modified random variable $V$ you are interested in (whatever it is) can be written uniquely as a ratio? $\endgroup$ – Did Sep 12 '13 at 6:08

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